Past exam of the condensed matter physics major option course of the University of Oxford / 2015 / 5 / b / Solution

Use the time dependence actually specified, $e^{i\omega t}$, and $\mathbf{B}=B\hat{\mathbf{z}}$. The component equations areAdding the second equation multiplied by $\pm i$ diagonalizes the response. With the literal definitions $v_{\pm }=v_x\pm i{v}_y$, $E_{\pm }=E_x\pm i{E}_y$, and $J_{\pm }=-ne{v}_{\pm }$,The source's printed $\omega \pm {\omega }_c$ reverses the helicity labels for these definitions. It is recovered by naming the two circular components oppositely, reversing $B$, or changing the matching time/polarization convention. As a concrete check, with $E_{+}\ne 0$, $E_{-}=0$ and positive $B$, the collisionless component equation resonates at $\omega ={\omega }_c$. The printed plus-labeled denominator would instead remain proportional to $2i{\omega }_c$. The physical result is unambiguous: only the circular polarization rotating with the electron orbit has positive-frequency cyclotron resonance.

For the convention used here, the dielectric eigenvalues are ${ϵ}_{\pm }={ϵ}_{\infty }+{\sigma }_{\pm }/(i{ϵ}_0\omega )$. In the lossless limit,These eigenvalues determine propagation and reflection in the Faraday geometry; changing the helicity names swaps their subscripts without changing any optical prediction.